Question

In $H{S}^{,}{I}^{,}R-N{H}_2,N{H}_3$ order of proton excepting tendency will be:

• A. $I^{-}>N{H}_3>RN{H}_2>H{S}^{-}$
• B. $N{H}_3>R-N{H}_2>H{S}^{-}>I^{-}$
• C. $RN{H}_2>N{H}_3>H{S}^{-}>I^{-}$
• D. $H{S}^{-}>RN{H}_2>N{H}_3>I^{-}$

Answer 1

The correct option is C $RN{H}_2>N{H}_3>H{S}^{-}>I^{-}$

Order will be $RN{H}_2>N{H}_3>H{S}^{-}>I^{-}$

Due to lone pair on $N^{-}$ atom which is further intensified by $R^{-}$. So, $RN{H}_2$ has stronger proton accepting than $N{H}_3$. Between $H{S}^{-},I^{-},N{H}_3$ as the size increases, proton accepting tendency decreases due to hindrance.